[RESOLVED] using DB for Select option ?

codegirl42

Hi all...

I'm trying to create a <select></select> field with data from a table.

It is a STATE table and the user needs to select the state and have the state NAME inserted into the user's table.

State table is setup like:

CREATE TABLE state (
state_id int(11) NOT NULL auto_increment,
country_id int(11) NOT NULL default '0',
name varchar(255) NOT NULL default '',
code varchar(10) default NULL,
PRIMARY KEY (state_id)
) TYPE=MyISAM AUTO_INCREMENT=85 ;

Query is setup like:

$user ="<select>
$query = 'SELECT * FROM state';
//Execute Query
$result = mysql_query($query);
$x = 0;
while ($x < mysql_num_rows($result)) :
$states = mysql_result($result, $x, 'name');
print "<option value='$name' ";
if (state==$states) print "selected";
print">$name</option>";
$x++;
endwhile;
print "</select>";

the select box shows up in the html...but none of the states from the DB are there....I'm stuck...please help..

Thank you

drew010

you have

$states = mysql_result($result, $x, 'name');
print "<option value='$name' ";
if (state==$states) print "selected";
print">$name</option>";

but $name doesnt appear to be set anywhere. you have probably mixed $name up with $states.

try it as

$states = mysql_result($result, $x, 'name');
print "<option value='$states' ";
if (state==$states) print "selected";
print">$states</option>";

codegirl42

okay, i put

$quert = 'SELECT * FROM state';
$result = mysql_query($query);
$x = 0;
while ($x < mysql_num_rows($result)) :
$states = mysql_result($result, $x, 'name');
print "<option value='$states' ";
if (state==$states) print "selected";
print">$states</option>";
$x++;
endwhile;

and i get this

Warning: mysql_result(): name not found in MySQL result index 6 in /home/urbanweb/public_html/members/users/index.php on line 43

drew010

try

echo '<select name="state">';
$query = "SELECT * FROM state";
$result = mysql_query($query);

while ($data = mysql_fetch_array($result)) {
  $states = $data['name'];
  echo "<option value=\"$states\"";
  if ($state == $states) echo " selected";
  echo ">$states</option>";
}
echo "</select>\n";

codegirl42

Perfect! Thank You!

codegirl42

Okay, I had to tweak it to fit with my layout...
but, for some reason, the option that they've ALREADY selected isnt chosen when they load the page....please help. 🙁

$user.="<tr><td>State:</td><td><select name='state'>";
$query = "SELECT * FROM state";
$result = mysql_query($query);
while ($data = mysql_fetch_array($result)) {
$states = $data['name'];
$user.="<option value='$states'";
if ($state == $states)
$user.=" selected";
$user.=">$states</option>";
}
$user.="</select></td></tr>";

drew010

where do they select an option from and what is it called?
if ($state == $states)
$user.=" selected";

its populated from that, most likely $state isnt the name of the variable.

codegirl42

$user.="<option value='$states'";

They can select and it IS sent to the Database....but when the form reloads, it goes back to the top of the select list INSTEAD of what they selected...

leatherback

So..

Is the keyword selected echoed out properly? If not, check what is matched against what. Echo out the relevant variables at locations which are crucial, such as the if(.. == .. )

You should use {} after your if statement, for consistency

codegirl42

okay....figured this one out...

if ($row["state"] == $states) {
$user.=" selected";
}

thanks guys...

leatherback

If it is solved, could you please mark the thread resolved (the\read tools on top)

codegirl42

sorry...just gettin the hang of all of this.. 🙂